PROBLEM II. 2. From a given point without a straight line to draw a perpendicular to that line. Let C be the point and A B the line. cutting AB in two points E and F; with E and F as centres, with a radius greater A- PROBLEM III. 3. From a given point in a straight line to erect a perpendicu lar to that line. Let C be the given point and A B the given line. With C as a centre describe an arc cutting A B in D and E; with D and E B C D E as centres, with a radius greater than 4- Second Method. With C as a centre describe an arc DEF; take the distances DE and EF equal to CD, and from E and as centres, with a radius greater than half the distance from E to F, describe arcs intersecting at G. Draw CG, and it is the perpendicular required (III. 33; III. 16; III. 15). Α Ꭰ E F -B C Draw the с 1st. Let A B be the given arc. chord AB and bisect it with a perpendicular (1; III. 16). 2d. Let C be the given angle. With Cas a centre describe an arc cutting the sides of the angle in A and B; bisect the arc A B with the line CD, and it will also bisect the angle C (III. 11). D PROBLEM V. 5. At a given point in a straight line to make an angle equal. to a given angle. Let A be the given point in the line AB, and C the given angle. With C as a centre describe an arc DE cutting the sides of the angle C; with A as a centre, with the same radius, describe an arc; with F as a centre, with a radius equal to the distance from D to E, describe an arc cutting the A G F E D arc FG. Draw A G. The angle A = C (III. 12; III. 11). PROBLEM VI. 6. Through a given point to draw a line parallel to a given straight line. 7. Two angles of a triangle given, to find the third. 8. The three sides of a triangle given, to construct the triangle. Take A B equal to one of the given sides; with A as a centre, with a radius equal to another of the given sides, describe an arc, and with B as a centre, with a radius equal to the remaining side, describe an arc intersecting the first arc at C. evidently the triangle required. Draw A C and CB, and ACB is PROBLEM IX. 9. Two sides and the included angle of a triangle given, to construct the triangle. Draw A B equal to one of the given sides; at B make the angle ABC equal to the given angle (5), and take BC equal to the other given side; join A and C, and A B C is evidently the triangle required. C A B PROBLEM X. 10. Two angles and a side of a triangle given, to construct the triangle. C If the angles given are not both adjacent to the given side, find the third angle by (7). Then draw A B equal to the given side, and at B make an angle ABC equal to one of the angles adjacent to A B, and at A make an angle B A C equal to the other angle adjacent to A B, and A B C is evidently the triangle required. A4 B PROBLEM XI. 11. Two sides of a triangle and the angle opposite one of them given, to construct the triangle. Draw an indefinite line AC; at A make the angle CAB equal to the given angle, and take A B equal to the side adjacent to the given angle; with A D B B as a centre, with a radius equal to the other given side, describe an arc cutting A C. If the given angle A is acute, B 1st. The given side BC, opposite the given angle, may be less than the other given side; then the arc described from B as a centre will cut AC in two points, C and D, A on the same side of A, and, drawing BC and BD, the triangles ABC and ABD (whose angle BDA is the supplement of the angle BCA), both satisfy the given conditions. 2d. The given side opposite the given angle may be equal to the perpendicular BE; then the arc described from B as a centre will touch A C, and the right triangle A BE is the only one that can satisfy the given conditions. 3d. The side opposite the given angle may be greater than the other given side; then the arc described from B as a centre AC will cut AB in C, and in another point on the other side of A. In this case there can be but one triangle A B C satisfying the given conditions, the triangle formed on the opposite side of A B containing not the given angle but its supplement. 4th. If the given angle is obtuse, the given side opposite the given angle must be greater than the other given side, and as in the last case above there can be but one solution. 12. Scholium. If the side opposite the given angle A is less than the perpendicular, or if the given angle is right or obtuse, and at the same time the side opposite the given angle is less than the other given side, the solution is impossible. 13. Corollary. From this and the preceding Problem and Theorems VIII., IX., and XIV. of Book I., it follows that with the exception of the ambiguity pointed out in the first part of this Problem, two triangles are equal if any three parts, of which one is a side, of the one arc equal to the corresponding parts of the other. |